RD Chapter 21 Some Special Series Ex 21.2 Solutions
Question - 1 : - Sum the following series to n terms:3 + 5 + 9 + 15 + 23 + ………….
Answer - 1 : -
Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.
We have,
Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn … (1)
Equation (1) can berewritten as:
Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2)from (1) we get
Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn
Sn = 3+ 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn
0 = 3 + [2 + 4 + 6 + 8+ … + (Tn – Tn-1)] – Tn
The difference betweenthe successive terms are 5-3 = 2, 9-5 = 4, 15-9 = 6,
So these differencesare in A.P
Now,
∴ The sum of the seriesis n/3 (n2 + 8)
Question - 2 : - 2 + 5 + 10 + 17 + 26 + ………..
Answer - 2 : -
Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.
We have,
Sn = 2+ 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn … (1)
Equation (1) can berewritten as:
Sn = 2+ 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2)from (1) we get
Sn = 2+ 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn
Sn = 2+ 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn
0 = 2 + [3 + 5 + 7 + 9+ … + (Tn – Tn-1)] – Tn
The difference betweenthe successive terms are 3, 5, 7, 9
So these differencesare in A.P
Now,
∴ The sum of the seriesis n/6 (2n2 + 3n + 7)
1 + 3 + 7 + 13 + 21 + …
Answer - 3 : -
Let Tn bethe nth term and Sn be the sum to n terms of the given series.
We have,
Sn = 1+ 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn … (1)
Equation (1) can berewritten as:
Sn = 1+ 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2)from (1) we get
Sn = 1+ 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn
Sn = 1+ 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn
0 = 1 + [2 + 4 + 6 + 8+ … + (Tn – Tn-1)] – Tn
The difference betweenthe successive terms are 2, 4, 6, 8
So these differencesare in A.P
Now,
∴ The sum of the seriesis n/3 (n2 + 2)
3 + 7 + 14 + 24 + 37 + …
Answer - 4 : -
Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.
We have,
Sn = 3+ 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn … (1)
Equation (1) can berewritten as:
Sn = 3+ 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2)from (1) we get
Sn = 3+ 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn
Sn = 3+ 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn
0 = 3 + [4 + 7 + 10 +13 + … + (Tn – Tn-1)] – Tn
The difference betweenthe successive terms are 4, 7, 10, 13
So these differencesare in A.P
Now,
∴ The sum of the seriesis n/2 [n2 + n + 4]
Question - 5 : - 1 + 3 + 6 + 10 + 15 + …
Answer - 5 : -
Let Tn bethe nth term and Sn be the sum to n terms of the givenseries.
We have,
Sn = 1+ 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn … (1)
Equation (1) can berewritten as:
Sn = 1+ 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2)from (1) we get
Sn = 1+ 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn
Sn = 1+ 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn
0 = 1 + [2 + 3 + 4 + 5+ … + (Tn – Tn-1)] – Tn
The difference betweenthe successive terms are 2, 3, 4, 5
So these differencesare in A.P
Now,
∴ The sum of the seriesis n/6 (n+1) (n+2)